Find the value of a convergent series

Question

So the series $$\frac{\pi}{2}-\frac{\pi^{3}}{8 \cdot 3!} + \frac{\pi^{5}}{32 \cdot 5!} - \cdot\cdot\cdot$$ is convergent, I need to find the value where it converges to

So far I have determined the pattern should be an alternating series $$\sum_{n=1}^{\infty} (-1)^{n+1} \, \frac{\pi^{2n-1}}{2^{2n-1} \cdot (2n-1)!}$$

But I need some help finding the value of convergence

Look up the Taylor expansion of the $sin$ function , the answer should be 1 — Mohammed M. Zerrak, Apr 29 '21 at 07:42

score 1 · Accepted Answer · answered Apr 29 '21 at 07:54

1

$(-1)^{n+1}=(-1)^{n-1}=i^{2n-2}$

$$\sum_{r=1}^\infty\dfrac{\left(\dfrac{i\pi}2\right)^{2n-1}}{i(2n-1)!}=\dfrac{e^{\dfrac{i\pi}2}-e^{\dfrac{-i\pi}2}}{2i}=\sin\dfrac{\pi}2$$

using $\displaystyle e^x=\sum_{r=0}^\infty\dfrac{x^r}{r!}$

and Simple Proof of the Euler Identity $\exp{i\theta}=\cos{\theta}+i\sin{\theta}$

answered Apr 29 '21 at 07:54

lab bhattacharjee

274,582

Hi there, my bad. It's actually that ${ a_{n} }$ converges to $\frac{1}{2}$ – Sirou Ewei Apr 29 '21 at 17:40

Find the value of a convergent series

1 Answers1